Chapter 7 - Applications of Integration - 7.6 Exercises - Page 494: 17

$$\eqalign{ & m = \frac{1}{{12}}\rho \cr & {M_x} = \frac{1}{{35}}\rho \cr & {M_y} = \frac{1}{{20}}\rho \cr & \left( {\overline x ,\overline y } \right) = \left( {\frac{3}{5},\frac{{12}}{{35}}} \right) \cr} $$

$$\eqalign{ & y = {x^2},{\text{ }}y = {x^3} \cr & {x^2} = {x^3} \cr & {x^2} - {x^3} = 0 \cr & {x^2}\left( {x - 1} \right) = 0 \cr & x = 0,{\text{ }}x = 1 \cr & {x^2} \geqslant {x^3}{\text{ on the interval }}\left[ {0,1} \right],{\text{ then}} \cr & {\text{Let }}f\left( x \right) = {x^2}{\text{ and }}g\left( x \right) = 0,{\text{ on the interval }}\left[ {0,1} \right] \cr & \cr & {\text{*The mass of the lamina is }} \cr & m = \rho \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr & m = \rho \int_0^1 {\left( {{x^2} - {x^3}} \right)} dx \cr & m = \rho \left[ {\frac{1}{3}{x^3} - \frac{1}{4}{x^4}} \right]_0^1 \cr & m = \rho \left[ {\frac{1}{3}{{\left( 1 \right)}^3} - \frac{1}{4}{{\left( 1 \right)}^3}} \right] - \rho \left[ {\frac{1}{3}{{\left( 0 \right)}^3} - \frac{1}{4}{{\left( 0 \right)}^3}} \right] \cr & m = \frac{1}{{12}}\rho \cr & \cr & *{\text{The moment about the }}x{\text{ - axis is}} \cr & {M_x} = \rho \int_a^b {\left[ {\frac{{f\left( x \right) + g\left( x \right)}}{2}} \right]} \left[ {f\left( x \right) - g\left( x \right)} \right]dx \cr & {M_x} = \rho \int_0^1 {\left[ {\frac{{{x^2} - {x^3}}}{2}} \right]} \left[ {{x^2} + {x^3}} \right]dx \cr & {M_x} = \frac{1}{2}\rho \int_0^1 {\left( {{x^4} - {x^6}} \right)} dx \cr & {M_x} = \frac{1}{2}\rho \left[ {\frac{1}{5}{x^5} - \frac{1}{7}{x^7}} \right]_0^1 \cr & {M_x} = \frac{1}{2}\rho \left[ {\frac{1}{5}{{\left( 1 \right)}^5} - \frac{1}{7}{{\left( 1 \right)}^7}} \right] - \frac{1}{2}\rho \left[ 0 \right] \cr & {M_x} = \frac{1}{{35}}\rho \cr & \cr & *{\text{The moment about the }}y{\text{ - axis is}} \cr & {M_y} = \rho \int_a^b x \left[ {f\left( x \right) - g\left( x \right)} \right]dx \cr & {M_y} = \rho \int_0^1 x \left[ {{x^2} - {x^3}} \right]dx \cr & {M_y} = \rho \int_0^1 {\left( {{x^3} - {x^4}} \right)} dx \cr & {M_y} = \rho \left[ {\frac{1}{4}{x^4} - \frac{1}{5}{x^5}} \right]_0^1 \cr & {M_y} = \rho \left[ {\frac{1}{4}{{\left( 1 \right)}^4} - \frac{1}{5}{{\left( 1 \right)}^5}} \right] - \rho \left[ {\frac{1}{4}{{\left( 0 \right)}^4} - \frac{1}{5}{{\left( 0 \right)}^5}} \right] \cr & {M_y} = \frac{1}{{20}}\rho \cr & \cr & *{\text{The center of mass }}\left( {\overline x ,\overline y } \right){\text{ is given by:}} \cr & \overline x = \frac{{{M_y}}}{m} = \frac{{\frac{1}{{20}}\rho }}{{\frac{1}{{12}}\rho }} = \frac{3}{5} \cr & \overline y = \frac{{{M_x}}}{m} = \frac{{\frac{1}{{35}}\rho }}{{\frac{1}{{12}}\rho }} = \frac{{12}}{{35}} \cr & \left( {\overline x ,\overline y } \right) = \left( {\frac{3}{5},\frac{{12}}{{35}}} \right) \cr & \cr & {\text{Summary}} \cr & m = \frac{1}{{12}}\rho \cr & {M_x} = \frac{1}{{35}}\rho \cr & {M_y} = \frac{1}{{20}}\rho \cr & \left( {\overline x ,\overline y } \right) = \left( {\frac{3}{5},\frac{{12}}{{35}}} \right) \cr} $$